Answer
$$
\int_{1}^{4} \sqrt{1+\left[f^{\prime}(x)\right]^{2}} d x \geq 3
$$
Work Step by Step
Since
$f^{\prime}(x)^{2} \geq 0,$
we know that
$\sqrt{1+\left[f^{\prime}(x)\right]^{2}} \geq \sqrt{1}=1.$
Then the arc length of the graph of $f(x)$ on $[1,4]$ is
$$
\int_{1}^{4} \sqrt{1+\left[f^{\prime}(x)\right]^{2}} d x \geq \int_{1}^{4} 1 d x=3
$$